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An Integral Representation for the Riemann Zeta Function on Positive Integer Arguments

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An Integral Representation for the Riemann Zeta Function on Positive Integer Arguments An integral representation for the Riemann zeta function on positive integer arguments Sumit Kumar Jha IIIT-Hyderabad, India Email: [email protected] May 23, 2019 Abstract In this brief note, we give an integral representation for the Riemann zeta function for positive integer arguments. To the best of our knowledge, the representation is new. Keywords: Ramanujan’s Master Theorem; Riemann Zeta Function; Polygamma Function; Polylogarithm Function AMS Classification: 33E20 We prove the following Theorem 1. We have, for integers r ≥ 2, Z ¥ −1=2 1 x Lir(−x) z(r) = r dx (1) p(2 − 2 ) 0 1 + x ¥ xk ¥ 1 where Lir(x) = ∑k=1 kr is the polylogarithm function, and z(r) = ∑n=1 nr is the Riemann zeta function. The above can be obtained as a direct consequence of the following result Theorem 2. For integers r ≥ 2 and 0 < n < 1, we have Z ¥ Li (−x) p y − (1 − n) xn−1 r dx = z(r) + (−1)r−1 r 1 .( 2) 0 1 + x sin np (r − 1)! ¥ xk dn G0(x) where Lir(x) = ∑k=1 kr is the polylogarithm function, yn(x) = dxn G(x) is the Polygamma function, G(x) is the Gamma function, and z(r) is the Riemann zeta function. Proof. Let n (r) 1 Hn = ∑ r k=1 k be the generalized Harmonic number. Then, we have the following generat- ing function from [2] Li (x) ¥ r = H(r)xn, − ∑ n 1 x n=1 for jxj< 1. We can also write ¥ Lir(−x) (r) n = ∑ Hn (−x) .( 3) (1 + x) n=1 We have for the following explicit form from [3] y − (n + 1) H(r) = z(r) + (−1)r−1 r 1 .( 4) n (r − 1)! 1 2 (r) r−1 yr−1(1) First note in this form, we have, H0 = z(r) + (−1) (r−1)! = 0. Next, we use Ramanujan’s Master Theorem (RMT) from [1], which is, Z ¥ p xs−1ff(0) − xf(1) + x2f(2) − · · ·g dx = f(−s),( 5) 0 sin sp where the integral is convergent for 0 < Re(s) < 1, and after certain condi- tions are satisfied by f. Now, using RMT with equations (3) and (4) gives us required equation (2). Proof of Theorem 1. We get the required equation (1) by plugging in n = 1=2 in equation (2), and using the following fact from [3] 1 r r y − = (−1) · (r − 1)! ·(2 − 1) · z(r). r 1 2 references 1. Amdeberhan, T., Espinosa, O., Gonzalez, I., Harrison, M., Moll, V. H., & Straub, A. (2012). Ramanujan’s master theorem. The Ramanujan Journal, 29(1-3), 103-120. 2. Weisstein, Eric W. ”Polylogarithm.” From MathWorld–A Wolfram Web Resource. 3. Weisstein, Eric W. ”Polygamma Function.” From MathWorld–A Wol- fram Web Resource.
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